Probably 2 very easy probability problems

Smifsret · #1 $Old$ May 13th, 2008, 03:30 PM

Hi, I need help understanding 2 probability problems.

I have a box with 24 circuit boards and I know 5 are faulty, if I take out 6 boards what is the probability that one of those 6 is faulty?

I am unsure about the second one althought it seems very easy, almost to easy.

A medicine makes 0.07 sick, what is the probability that 3 out of 10 people that take it will get sick?

I have a hard time following the examples in the book so I hope you can shed some light on what seems to be 2 easy questions.

Thanks!

galactus · #2 $Old$ May 13th, 2008, 03:52 PM

The second one is a binomial probability.

$C(10,3)(\frac{7}{100})^{3}(\frac{93}{100})^{7}$

galactus · #3 $Old$ May 13th, 2008, 03:55 PM

The first one is a hypergeometric probability,

You choose 1 from the 5 faulty ones and 5 from the 19 good ones.

$\frac{C(5,1)C(19,5)}{C(24,6)}$

Smifsret · #4 $Old$ May 13th, 2008, 04:30 PM

Thank you but can you explain how to proceed from there?

galactus · #5 $Old$ May 13th, 2008, 05:01 PM

Run it through a calculator. C(10,3), for instance, means the number ways to choose 3 items out of 10. C(10,3)=120

The formula is $\frac{n!}{(n-k)!k!}=\frac{10!}{(10-3)!3!}=120$

Smifsret · #6 $Old$ May 13th, 2008, 05:26 PM

Ok now I feel even dumber, when I run that I get 4320.

Also, how do i get the % of 3 out of 10 if I even get 120?

Sorry to be such a bother.

Smifsret · #7 $Old$ May 15th, 2008, 03:09 AM

Can I get these answers checked please?

Question 1 = 5*19*18*17*16*15*6*5*4*3*2*1 / (1*5*4*3*2*1*24*23*22*21*20*19*18) = 0.02400 = 2.4%

Question 2 = 10*9*8/3/2/1*0.07^3 * (1-0.07)^7 = 0.02477 = 2.5%

Thanks.